Theoretical Yield as a MASS


On this screen you will see how to calculate the theoretical yield of a product when the product is measured as a MASS.



Consider the following reaction:
CH3COOH + C2H5OH ---------» «--------- CH3COOC2H5 + H20
The reaction involves 30 grams of Ethanoic Acid and 50 grams of Ethanol.
How do you calculate the theoretical yield in grams of Ethyl Ethanoate?
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Find which of the reactants CH3COOH + C2H5OH is the limiting reactant.
The limiting reactant is the one which would produce the lesser amount of product.
a Find the number of moles of each reactant involved, where Moles = Mass/RMM.
Moles of Ethanoic Acid   = 30/60.0       = 0.5 moles          RMM = 60.0
Moles of Ethanol            = 50/46.0       = 1.087 moles       RMM = 46.0
b Find how many moles of product could be produced from each reactant.
From the balanced equation, you can see that:
1 mole of Ethanoic Acid would produce 1 mole of Ethyl Ethanoate. Thus:
0.5 moles of Ethanoic Acid would produce 0.5 moles of Ethyl Ethanoate.
1 mole of Ethanol would produce 1 mole of Ethyl Ethanoate. Thus:
1.087 moles of Ethanol would produce 1.087 moles of Ethyl Ethanoate.
Ethanoic Acid would produce less Ethyl Ethanoate and is therefore the limiting reactant.
Find the Mass of Ethyl Ethanoate that can be produced.
From Step 1 you know that 0.5 moles of Ethyl Ethanoate can be produced.
Thus Mass of Ethyl Ethanoate = Moles x RMM                    RMM = 88.0
Thus Mass of Ethyl Ethanoate = 0.5 x 88.0 g = 44.0 g
Theoretical Yield of Ethyl Ethanoate is 44.0 grams.